Solve 3ax^2-9ax-12a=

$$3 a x ^ { 2 } - 9 a x - 12 a =$$

Factor

$$3a\left(x-4\right)\left(x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(ax^{2}-3ax-4a\right)$$

Consider ax^{2}-3ax-4a. Factor out a.

$$a\left(x^{2}-3x-4\right)$$

Consider x^{2}-3x-4. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+px+qx-4. To find p and q, set up a system to be solved.

$$p+q=-3 pq=1\left(-4\right)=-4$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$p=-4 q=1$$

Rewrite x^{2}-3x-4 as \left(x^{2}-4x\right)+\left(x-4\right).

$$\left(x^{2}-4x\right)+\left(x-4\right)$$

Factor out x in x^{2}-4x.

$$x\left(x-4\right)+x-4$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$3a\left(x-4\right)\left(x+1\right)$$

Evaluate

$$3a\left(x-4\right)\left(x+1\right)$$